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Homework Help: Solve for constants a and b given the limit of the function

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine all the numbers a and b for which the limit of the function as x approaches 2 equals to 4

    sorry, im kinda new at this, not sure how to type it out
     

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  2. jcsd
  3. Sep 26, 2010 #2

    fzero

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    Does writing

    [tex]\sqrt{ax +b} -3 = \sqrt{a(x-2) + b+2a} -3[/tex]

    give you any ideas?
     
  4. Sep 26, 2010 #3
    try multiply by its Conjugate that will help you
     
  5. Sep 26, 2010 #4
    Oh, thank you
    Ive gotten to a point where (x-2) cancels out, and i can sub in x=2 without the denominator coming out as undefined..
    But i'm still not sure where to go from here
     
  6. Sep 26, 2010 #5

    fzero

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    Can you explain a bit more of your work? If you're using my hint then you expanded the square root in powers of (x-2) and there's a divergent term, a term that's independent of x, and then terms that vanish as x -> 2. The coefficients of the first 2 terms can be thought of as a pair of equations to solve for a and b once you demand that the divergent part vanishes and that the finite term gives the correct limit.
     
  7. Sep 28, 2010 #6

    Hi fzero, how did you get that second part?
    Thanks!
     
  8. Sep 28, 2010 #7
    I'm not sure how sqr.rt.((x-2) + b + 2a) shows me how to get there from the original numerator. What did you multiply, or how did you substitute?
     
  9. Sep 28, 2010 #8

    fzero

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    All I did was add and subtract 2 a under the square root so that I could express the numerator in terms of the x-2 that appeared in the denominator. However, if you're not familiar with series expansions, the next step would be far from obvious.
     
  10. Sep 28, 2010 #9
    I figured out how you rewrote the numerator, and I was able to eliminate the (x-2) in the numerator and denominator. I am now left with (3a+b-3)/sqr.rt.(b+2a+3) = 4. Do you have any hints as to what comes after that?
     
  11. Sep 28, 2010 #10

    fzero

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    There's also a term proportional to 1/(x-2). This gives a 2nd equation to use to solve for a,b.
     
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